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# Data Structures and Algorithms | Set 14

• Difficulty Level : Medium
• Last Updated : 05 Jul, 2018

Following questions have been asked in GATE CS 2008 exam.

1. We have a binary heap on n elements and wish to insert n more elements (not necessarily one after another) into this heap. The total time required for this is
(A) Θ(logn)
(B) Θ(n)
(C) Θ(nlogn)
(D) Θ(n2)

The worst case time complexity for insertion in a binary heap is O(Logn) (Refer Wiki). So inserting n elements in a heap of size n should take Θ(nlogn) time.
But choice (B) seems to be more appropriate answer. One of the solution of O(n) complexity can be to take the ‘n’ elements of the heap and other ‘n’ elements together and construct heap in O(2n) = O(n). Thanks to pankaj for suggesting this solution.

2. The Breadth First Search algorithm has been implemented using the queue data structure. One possible order of visiting the nodes of the following graph is

(A) MNOPQR
(B) NQMPOR
(C) QMNPRO
(D) QMNPOR

Answer (C)

3. Consider the following functions:

```  f(n)   = 2^n
g(n)   = n!
h(n)   = n^logn ```

Which of the following statements about the asymptotic behaviour of f(n), g(n), and h(n) is true?
(A) f(n) = O(g(n)); g(n) = O(h(n))
(B) f(n) = Ω(g(n)); g(n) = O(h(n))
(C) g(n) = O(f(n)); h(n) = O(f(n))
(D) h(n) = O(f(n)); g(n) = Ω(f(n))

Answer (D)

According to order of growth: h(n) < f(n) < g(n) (g(n) is asymptotically greater than f(n) and f(n) is asymptotically greater than h(n) ) We can easily see above order by taking logs of the given 3 functions

`   lognlogn < n < log(n!)  (logs of the given f(n), g(n) and h(n)).`

Note that log(n!) = Θ(nlogn)

4. The minimum number of comparisons required to determine if an integer appears more than n/2 times in a sorted array of n integers is

(A) Θ(n)
(B) Θ(logn)
(C) Θ(log*n)
(D) Θ(n)

Answer (B)

Please see the post Check for Majority Element in a sorted array for details.

Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above.

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